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\title{Work2: 2.9.1 Theoretical questions}


\author{邵盛栋 \\ 信息与计算科学 3200103951}

\begin{document}
	\maketitle
	\section*{习题解答}
	\begin{enumerate}[I]
		\item \begin{itemize}
			\item 由于$ x_{0}=1,x_{1}=2,f(x)=\frac{1}{x} $,则$ p_{1}(f;x) $经过点$ (1,1),(2,\dfrac{1}{2}) $\\
			解得$ p_{1}(f;x)=-\frac{1}{2}x+\frac{3}{2} $\\
			则$ f''(\xi(x))=\dfrac{2(f(x)-p_{1}(f;x))}{(x-1)(x-2)}=\dfrac{1}{x} $,即$ \dfrac{3}{(\xi(x))^{3}}=\dfrac{1}{x} $\\
			所以$ \xi(x)=\sqrt[3]{2x} $.
			\item 将定义域延拓至$ [1,2] $,则$ max\xi(x)=\xi(2)=\sqrt[3]{4},min\xi(x)=\xi(1)=\sqrt[3]{2} $\\
			$ maxf''(\xi(x))=f''(\xi(1))=1 $.
		\end{itemize}
		\item 由定理2.5，由于$ f_{i}\geq0, \forall i=0,1,\ldots,n $,则对于给定 $x_0, x_1, \ldots, x_n \in \mathbb{C}$ 和对应的 $\sqrt{f_0}, \sqrt{f_1}, \ldots, \sqrt{f_n} \in \mathbb{C}$，存在一个唯一的多项式 $p_n(x) \in \mathbb{P}_n$ 使得
		\[\forall i=0,1, \ldots, n, \quad p_n\left(x_i\right)=\sqrt{f_i}\]
		令$ p(x_{i})=(p_{n}(x_{i}))^{2}=f_{i}\geq0 \quad \forall i=0,1, \ldots, n $,则$ p(x) $满足$ p(x) \in \mathbb{P}_{2n}^{+} $,\\
		此$ p(x) $即满足题中条件.
		\item \begin{itemize}
			\item \begin{proof}
				由数学归纳法，当 $ n=1 $时，$ f[t,t+1]=e^{t+1}-e^{t}=\dfrac{(e-1)^{1}}{1!}e^{t} $成立,\\
				假设当$ n=k $时，$ f[t,t+1,\ldots,t+k]=\dfrac{(e-1)^{k}}{k!}e^{t} $成立,\\
				当$ n=k+1 $时，由定理2.17，
				\begin{equation*}
					\begin{split}
						f[t,t+1,\ldots,t+k+1]&=\dfrac{f[t+1,t+2,\ldots,t+k+1]-f[t,t+1,\ldots,t+k]}{k+1}\\
						&=\dfrac{\frac{(e-1)^{k}}{k!}e^{t+1}-\frac{(e-1)^{k}}{k!}e^{t}}{k+1}\\
						&=\dfrac{(e-1)^{k+1}}{(k+1)!}e^{t}
					\end{split}
				\end{equation*}
			所以，$ \forall t\in\mathbb{R}\quad f[t,t+1,\ldots,t+n]=\dfrac{(e-1)^{n}}{n!}e^{t} $
			\end{proof}
			\item 令第一小题中的$ t=0 $，则$ f[0,1,\ldots,n]=\dfrac{(e-1)^{n}}{n!} $\\
			所以$ \exists \xi \in(0, n) \text { s.t. } f[0,1, \ldots, n]=\dfrac{(e-1)^{n}}{n!}=\dfrac{1}{n !} f^{(n)}(\xi) . $\\
			解得：$ \xi=n\ln(e-1) $,由于$ \ln(e-1)\approx0.54>\dfrac{1}{2} $\\
			所以$ \xi $位于中点$ \dfrac{n}{2} $的右侧.	
		\end{itemize}
		\item \begin{itemize}
			\item 根据定义2.18,可构造如下table of divided difference：\\
			\begin{table*}[ht]
				\centering
				\begin{tabular}{c|cccc}
					0&5&&&\\
					1&3&-2&&\\
					3&5&1&1&\\
					4&12&7&2&$ \frac{1}{4} $
				\end{tabular}
			\end{table*}
		
			根据定义2.14,插值多项式是由上表的主对角线和第一列生成，即：
			\[ p_{3}(f;x)=5-2x+x(x-1)+\dfrac{1}{4}x(x-1)(x-3)=\dfrac{1}{4}x^{3}-\dfrac{9}{4}x+5 \]
			\item 当$ p'_{3}(f;x)=\dfrac{3}{4}x^{2}-\dfrac{9}{4}=0 $时，$ x=\sqrt{3}(x\in(1,3)) $\\
			所以$ f $在$ x\in(1,3) $上去最小值时，对应的$ x_{min} $的近似值为$ \sqrt{3} $.
		\end{itemize}
		\item \begin{itemize}
			\item 由于$ f(x)=x^{7} $,则\\
			\[ f(0)=0,f(1)=1,f'(1)=7,f''(1)=42,f(2)=128,f'(2)=448 \]
			构造table of divided difference：\\
			\begin{table*}[ht]
				\centering
				\begin{tabular}{c|cccccc}
					0&0&&&&&\\
					1&1&1&&&&\\
					1&1&7&6&&&\\
					1&1&7&21&15&&\\
					2&128&127&120&99&42&\\
					2&128&448&321&201&102&30
				\end{tabular}
			\end{table*}
			
			因此，$ f[0,1,1,1,2,2]=30 $.
			\item 由定理2.37，$ \exists \xi \in(0, 2) \text { s.t. } \dfrac{f^{(5)}(\xi)}{5!}=30 $,\\
			即$ 21\xi^{2}=30 $,解得：$ \xi=\dfrac{\sqrt{70}}{7} $
		\end{itemize}
	\newpage
		\item \begin{itemize}
			\item 已知, $ f(0)=1,f(1)=2,f'(1)=-1,f(3)=f'(3)=0 $,构造table of divided difference：\\
			\begin{table*}[ht]
				\centering
				\begin{tabular}{c|ccccc}
					0&1&&&&\\
					1&2&1&&&\\
					1&2&-1&-2&&\\
					3&0&-1&0&$ \frac{2}{3}$& \\
					3&0&0&$ \frac{1}{2} $&$ \frac{1}{4} $&$ -\frac{5}{36} $
				\end{tabular}
			\end{table*}
			
			则$ p_{4}(f;x)=1+x-2x(x-1)+\dfrac{2}{3}x(x-1)^{2}-\dfrac{5}{36}x(x-1)^{2}(x-3) $\\
			所以, $ f(2)\approx p_{4}(f;2)=\dfrac{11}{18} $
			\item 由定理2.37, 
			\[ R_{4}(f;x)=\dfrac{f^{(5)}(\xi)}{5!}x(x-1)^{2}(x-3)^{2} \]
			所以，$ R_{4}(f;2)=\dfrac{f^{(5)}(\xi)}{60}\leq\dfrac{M}{60} $, 即最大可能误差为$ \dfrac{M}{60} $.
		\end{itemize}
		\item \begin{proof}
			由数学归纳法，\\
			当$ k=1 $时，$ \Delta f(x)=1 ! h^1 f[x_0, x_1]=(x_{1}-x_{0})\dfrac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}=f(x_{1})-f(x_{0}) $\\
			假设$ k=n $时，$ \Delta^n f(x)=n! h^n f[x_0, x_1, \ldots, x_n] $成立,\\
			当$ k=n+1 $时，
			\[ 	
			\begin{aligned}
				\Delta^{n+1} f(x) &=\Delta \Delta^n f(x)=\Delta^n f(x+h)-\Delta^n f(x)\\
				&=n! h^n f[x_1, x_2, \ldots, x_{n+1}]-n! h^n f[x_0, x_1, \ldots, x_{n}]\\
				&=n! h^n(f[x_1, x_2, \ldots, x_{n+1}]-f[x_0, x_1, \ldots, x_{n}])\\
				&=n! h^n(x_{n+1}-x_{0})f[x_0, x_1, \ldots, x_{n+1}]\\
				&=n! h^n((n+1)h)f[x_0, x_1, \ldots, x_{n+1}]\\
				&=(n+1)! h^{n+1}f[x_0, x_1, \ldots, x_{n+1}]
			\end{aligned} 
		    \]
		    所以，$ \Delta^k f(x)=k!h^k f[x_0, x_1, \ldots, x_k] $成立.
		\end{proof}
		同理，利用数学归纳法，可知$ \nabla^k f(x)=k ! h^k f[x_0, x_{-1}, \ldots, x_{-k}] $亦成立.
		\item 
		\begin{proof}
			\[
			\begin{aligned}
				\dfrac{\partial}{\partial x_{0}}f[x_0, x_1, \ldots, x_{n}]&=\lim\limits_{\Delta x_{0}\to0}\dfrac{f[x_0+\Delta x_{0}, x_1, \ldots, x_{n}]-f[x_0, x_1, \ldots, x_{n}]}{(x_{0}+\Delta x_{0})-x_{0}}\\
				&=\lim\limits_{\Delta x_{0}\to0}\dfrac{f[x_1,x_2, \ldots, x_{n},x_0+\Delta x_{0}]-f[x_0, x_1, \ldots, x_{n}]}{(x_{0}+\Delta x_{0})-x_{0}}\\
				&=\lim\limits_{\Delta x_{0}\to0}f[x_0,x_1, \ldots, x_{n},x_0+\Delta x_{0}]\\
				&=f[x_0,x_0,x_1, \ldots, x_{n}]
			\end{aligned}
			\]
		\end{proof}
		若是相对于其他变量$ (x_1,x_2,\ldots,x_n) $的偏导数，同理可得：
		\[ \dfrac{\partial}{\partial x_{k}}f[x_0, x_1, \ldots, x_{n}]=f[x_0,x_1, \ldots,x_{k-1},x_k,x_k,x_{k+1},\ldots,x_{n}] \quad k=1,2,\ldots,n\]
		\item $ \max\limits_{x\in[a,b]}|a_{0}x^{n}+a_{1}x^{n-1}+\dots+a_{n}|=|a_{0}|\max\limits_{x\in[a,b]}|x^{n}+\dfrac{a_{1}}{a_{0}}x^{n-1}+\dots+\dfrac{a_{n}}{a_{0}}| $\\
		由推论2.47，对于$ n\in\mathbb{N}^{+} $,
		\[ \max\limits_{x\in[-1,1]}|x^{n}+a_{1}x^{n-1}+\dots+a_{n}|\geq\dfrac{1}{2^{n-1}} \]
		则当$ x\in[a,b] $时，
		\[ 
		\max\limits_{x\in[a,b]}|x^{n}+a_{1}x^{n-1}+\dots+a_{n}|\geq\dfrac{1}{2^{n-1}}(\dfrac{b-a}{2})^{n}
		\]
		所以，
		\[
		\max\limits_{x\in[a,b]}|a_{0}x^{n}+a_{1}x^{n-1}+\dots+a_{n}|\geq\dfrac{|a_{0}|}{2^{n-1}}(\dfrac{b-a}{2})^{n}=\dfrac{|a_{0}|(b-a)^{n}}{2^{2n-1}}
		\]
		即$ \min \max\limits_{x\in[a,b]}|a_{0}x^{n}+a_{1}x^{n-1}+\dots+a_{n}|=\dfrac{|a_{0}|(b-a)^{n}}{2^{2n-1}} $
		\item \begin{proof}
			由定理2.44，$ |T_{n}(x)|\leq1 $且有定义$ x'_{k}=\cos \dfrac{k}{n}\pi,\quad k=0,1,\ldots,n $\\
			则$ \Vert\hat{p}_{n}(x)\Vert_{\infty}=\dfrac{1}{|T_{n}(a)|} $\\
			假设
			\[
			\exists p\in \mathbb{P}^{a}_{n}, \text{ s.t. } \Vert p\Vert_{\infty}<\dfrac{1}{|T_{n}(a)|}
			\]
			不妨设$ Q(x)=\dfrac{T_{n}(x)}{T_{n}(a)}-p(x) $,则
			\[
			Q(x'_{k})=\dfrac{(-1)^{k}}{T_{n}(a)}-p(x'_{k})
			\quad k=0,1,\ldots,n
			\]
			$ Q(x) $在$ (n+1) $个点上共有n次正负交替变换，则$ Q(x) $在$ [-1,1] $上共有n个零点.\\
			由于存在固定的$ a>1 $,使得$ p(a)=1 $,即$ Q(a)=0 $.\\
			所以$ Q(x)=0 $在$ \mathbb{R} $上至少有$ n+1 $个根，但$ Q(x) $的度数至多为$ n $，矛盾！\\
			所以，$ \forall p\in\mathbb{P}^{a}_{n},\quad \Vert\hat{p}_{n}\Vert_{\infty}\leq\Vert p\Vert_{\infty} $.
		\end{proof}
		\item \begin{proof}
			对于$ \forall k=0,1,\ldots,n,\forall t\in(0,1) $, $ b_{n,k}(t)>0 $显然成立.\\
			\[
			\begin{aligned}
				\sum_{k=0}^n b_{n, k}(t)&=\sum_{k=0}^n \left(\begin{array}{l}
					n \\
					k
				\end{array}\right) t^k(1-t)^{n-k}=(t+1-t)^{n}=1\\
				\sum_{k=0}^n k b_{n,k}(t)&=nt\sum_{k=0}^n \dfrac{k}{nt} b_{n,k}(t) \\
				&=nt\sum_{k=0}^n \left(\begin{array}{l}
					n-1 \\
					k-1
				\end{array}\right) t^{k-1}(1-t)^{n-k}\\
				&=nt(t+1-t)^{n-1}\\
				&=nt\\
				\sum_{k=0}^n(k-nt)^{2} b_{n,k}(t)&=n^{2}t^{2}\sum_{k=0}^n b_{n,k}(t)-2nt\sum_{k=0}^n k b_{n,k}(t)+\sum_{k=0}^n k^{2} b_{n,k}(t)\\
				&=n^{2}t^{2}-2nt(nt)+\sum_{k=0}^n k^{2} b_{n,k}(t)\\
				&=-n^{2}t^{2}+nt\sum_{k=0}^{n-1}(k+1)\left(\begin{array}{l}
					n-1 \\
					k
				\end{array}\right)t^{k}(1-t)^{n-1-t}\\
				&=-n^{2}t^{2}+nt(\sum_{k=0}^{n-1}k b_{n-1,k}(t)+\sum_{k=0}^{n-1}b_{n-1,k}(t))\\
				&=-n^{2}t^{2}+nt((n-1)t+1)\\
				&=nt(1-t)
			\end{aligned}
			\]		
		\end{proof}
	\end{enumerate}
	
\end{document}
